Question 19.2: Balancing a Redox Equation for a Reaction in Basic Solution ...

Step 1. Write the unbalanced net ionic equation.

ClO^{-} (aq) + Cr(OH)^{2-}_{4}(aq) → CrO^{2-}_{4}(aq) + Cl^{-}(aq)         Unbalanced

Step 2. Decide which atoms are oxidized and which are reduced, and write the two unbalanced half-reactions. The unbalanced net ionic equation shows that chromium is oxidized (from +3 to +6) and chlorine is reduced (from +1 to -1) Thus, we can write the following half-reactions:

Oxidation half-reaction          Cr(OH)^{-}_{4} (aq) → CrO^{2-}_{4} (aq)
Reduction half-reaction         ClO^{-} (aq) → Cl^{-}(aq)

Step 3. Balance both half-reactions for all atoms except O and H. The half- reactions are already balanced for atoms other than O and H.

Step 4. Balance each half-reaction for O by adding H_{2}O to the side with less O, and balance for H by adding H^{+} to the side with less H.

Oxidation   Cr(OH)^{-}_{4}(aq) → CrO^{2-}_{4}(aq) + 4 H^{+}(aq)
Reduction    ClO^{-}(aq) + 2 H^{+}(aq) → Cl^{-}(aq) + H_{2}O(l)

Step 5. Balance both half-reactions for charge by adding electrons (e ^{-} ) to the side with the greater positive charge.

Oxidation    Cr(OH)^{-}_{4}(aq) → CrO^{2-}_{4}(aq) + 4 H^{+}(aq)+3 e^{-}

Reduction     ClO^{-}(aq) + 2 H^{+}(aq) +2 e^{-} → Cl^{-}(aq) + H_{2}O(l)

Step 6. Multiply the half-reactions by integers to make the electron count the same in both half-reactions. The oxidation half-reaction must be multiplied by 2, and the reduction half-reaction must be multiplied by 3 to give 6 e^{-} in both:

Oxidation    2 × [Cr(OH)]^{-}_{4}(aq) → CrO^{2-}_{4}(aq) + 4 H^{+}(aq) + 3 e^{-}]
or                 2 Cr(OH)^{-}_{4}(aq) →  2 CrO^{2-}_{4}(aq) + 8 H^{+}(aq) + 6e^{-}

Reduction   3 ×  [ClO^{-}(aq) + 2 H^{+}(aq) + 2 e^{-} →   Cl^{-}(aq) + H_{2}O(l)]
or                 3 ClO^{-}(aq) + 6 H^{+}(aq) + 6 e^{-} →   Cl^{-}(aq) + 3 H_{2}O(l)

Step 7. Add the two half-reactions together, and cancel electrons and other species that occur on both sides of the equation. The electrons must always cancel.

Now, cancel the other species that appear on both sides of the equation. Since there are 8 H^{+} in the products and 6 H^{+} in the reactants, the result after cancellation is 2 H^{+}

Finally, since we know that the reaction takes place in basic solution, we must add 2 OH^{-} ions to both sides of the equation to neutralize the 2 H^{+} ions on the right, giving 2 additional H_{2}O.

The final net ionic equation, balanced for both atoms and charge, is